You don’t need calculus to do this. Neither one is accelerating, so “5 seconds after they started moving” is irrelevant. Just calculate the velocity of one in the reference frame of the other by subtracting the vectors: from the point of view of the boy, the girl’s velocity vector has orthogonal components of -5 ft/sec north and 1 ft/sec east, so the magnitude is 26^0.5 ft/sec.
The “5 seconds after they started moving” is relevant. If we assume this takes place on Earth (i.e. on the surface of a sphere with a set pair of north/south poles), the angle between the two vectors changes depending on their current position.
If it’s not on the equator, it’s also slightly up to interpretation if “Due East” means they’ll turn to stay on the same latitude, always adjusting to stay moving east forever or if they’ll do a great circle. In the former case, the north moving one will eventually get stuck at the north-pole too instead of continuing their circle around the globe. Most likely not within 5 seconds though, unless the place they started was within 25 feet of the north-pole.
To actually do the math we’ll need to know (or somehow deduce) where “the place where everything about them began” is though.
I guess the calculus portion of this is to write the separation as a function of time, s = √26*t, and then realize that the rate of separation is the same regardless of time, because the first derivative is a constant.
I mean, there’s quite a few calculus problems that you can do without calculus. Pretty much 80% of the optimization problems/max’s and minimums in a typical Calc 1 class can be done if you remember that -b/2a is the vertex of a quadratic.
The boy’s speed is given as 5 ft/sec, but the question is ambiguous as to whether his position remains due north of the starting point, or due north of the girl. Your approach assumes the former, but his 5ft/sec speed may include the girl’s 1ft/sec eastward component.
Technically, his starting location is defined contradictorily, being both the same position as hers, as well as “due north”. We can only proceed if a point can be considered “due north” of itself.
Technically, his initial direction of travel isn’t actually defined, nor is that direction specified as constant. Only his initial position and “5ft/sec” speed has actually been defined. The problem doesn’t actually constrain him to a northerly heading, or even to a constant heading. He could choose to orbit her at a constant distance as she travels eastward.
His possible position will be up to 25ft from the origin, putting him at most 30 feet from her, if his direction is opposite hers. If his direction is constant, his minimum distance from her will be 20 feet.
If his direction is not constant, and he elects to minimize his distance from her, as his distance from her approaches zero, his revolutions around her in a given time will approach infinite, and we will have to consider relativistic effects. From his perspective, his body will be ripped apart into a dizzying pink mist, which he will experience for all eternity. Poetic, I suppose.
You don’t need calculus to do this. Neither one is accelerating, so “5 seconds after they started moving” is irrelevant. Just calculate the velocity of one in the reference frame of the other by subtracting the vectors: from the point of view of the boy, the girl’s velocity vector has orthogonal components of -5 ft/sec north and 1 ft/sec east, so the magnitude is 26^0.5 ft/sec.
The “5 seconds after they started moving” is relevant. If we assume this takes place on Earth (i.e. on the surface of a sphere with a set pair of north/south poles), the angle between the two vectors changes depending on their current position.
If it’s not on the equator, it’s also slightly up to interpretation if “Due East” means they’ll turn to stay on the same latitude, always adjusting to stay moving east forever or if they’ll do a great circle. In the former case, the north moving one will eventually get stuck at the north-pole too instead of continuing their circle around the globe. Most likely not within 5 seconds though, unless the place they started was within 25 feet of the north-pole.
To actually do the math we’ll need to know (or somehow deduce) where “the place where everything about them began” is though.
What a convoluted way of asking the teacher to spill the beans. I like it.
I guess the calculus portion of this is to write the separation as a function of time, s = √26*t, and then realize that the rate of separation is the same regardless of time, because the first derivative is a constant.
I mean, there’s quite a few calculus problems that you can do without calculus. Pretty much 80% of the optimization problems/max’s and minimums in a typical Calc 1 class can be done if you remember that -b/2a is the vertex of a quadratic.
The boy’s speed is given as 5 ft/sec, but the question is ambiguous as to whether his position remains due north of the starting point, or due north of the girl. Your approach assumes the former, but his 5ft/sec speed may include the girl’s 1ft/sec eastward component.
The implication that the boy is strafing while maintaining a straight vertical line to the girl is hilarious
Technically, his starting location is defined contradictorily, being both the same position as hers, as well as “due north”. We can only proceed if a point can be considered “due north” of itself.
Technically, his initial direction of travel isn’t actually defined, nor is that direction specified as constant. Only his initial position and “5ft/sec” speed has actually been defined. The problem doesn’t actually constrain him to a northerly heading, or even to a constant heading. He could choose to orbit her at a constant distance as she travels eastward.
His possible position will be up to 25ft from the origin, putting him at most 30 feet from her, if his direction is opposite hers. If his direction is constant, his minimum distance from her will be 20 feet.
If his direction is not constant, and he elects to minimize his distance from her, as his distance from her approaches zero, his revolutions around her in a given time will approach infinite, and we will have to consider relativistic effects. From his perspective, his body will be ripped apart into a dizzying pink mist, which he will experience for all eternity. Poetic, I suppose.